Question

Solve the following differential equation:

`("x + y") "dy"/"dx" = 1`

Answer 1

`("x + y") "dy"/"dx" = 1`

∴ `"dx"/"dy" = "x + y"`

∴ `"dx"/"dy" - "x" = "y"`

∴ `"dx"/"dy" + (- 1)"x" = "y"`      ....(1)

This is the linear differential equation of the form

`"dx"/"dy" + "P"*"x" = "Q",` where P = - 1 and Q = y

∴ I.F. = `"e"^(int "P dy") = "e"^(int - 1 "dy") = "e"^-"y"`

∴ the solution of (1) is given by

x.(I.F.) = `int "Q" * ("I.F.") "dy" + "c"`

∴ `"x" * "e"^-"y" = int "y" * "e"^-"y" "dy" + "c"`

∴ `"e"^-"y" * "x" = "y" int "e"^-"y" "dy" - int ["d"/"dx" ("y") int "e"^-"y" "dy"] "dy" + "c"`

`= "y" * ("e"^-"y")/-1 - int 1 * ("e"^-"y")/-1 "dy" + "c"`

`= - "ye"^-"y" + int "e"^-"y" "dy" + "c"`

∴ `"e"^-"y" * "x" = - "ye"^-"y" + "e"^-"y"/-1 + "c"`

∴ `"e"^-"y" * "x" + "ye"^-"y" + "e"^-"y" = "c"`

∴ `"e"^"-y" ("x + y + 1") = "c"`

∴ x + y + 1 = ce^y 

This is the general solution.